An experimenter would like to construct a 99% two-sided t-interval, with a length at almost 0.2 ohms, for the average resistance of a segment of copper cable of a certain length. If the experimenter feels that the standard deviation of such resistances is no larger than 0.15 ohms, what sample size would you recommend? Assume that an appropriate t-value is 3.0.A. Aprox 10B. Aprox 20C. Aprox 30D. Aprox 40E. Aprox 50

Answer:Aprox 20Step-by-step explanation:A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". [tex]\bar X[/tex] represent the sample mean for the sample  [tex]\mu[/tex] population means represent the sample standard deviation n represent the sample size (variable of interest) Confidence =99% or 0.99The margin of error is given by this formula: [tex] ME=t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]    (1) And on this case we have that ME =0.2/2=0.1 ohms, since the lenght of the confidence interval is L=2ME and we are interested in order to find the value of n, if we solve n from equation (4) we got: [tex]n=(\frac{t_{\alpha/2} s}{ME})^2[/tex]   (2) The critical value for 99% of confidence interval is provided, [tex]t_{\alpha/2}=3.0[/tex], replacing into formula (2) we got: [tex]n=(\frac{3.0(0.15)}{0.1})^2 =20.25 \approx 21[/tex] So the answer for this case would be n=21 rounded up to the nearest integer And with the options provided the best answer is:Aprox 20